# Is the composite of splitting field and random field also Galois over base field?

I'm having trouble with a seemingly simple statement.

Suppose $$N|K$$ is a Galois extension with $$L$$ an additional field over $$K$$. Now for $$\sigma \in \text{Gal}([N,L] | L)$$, with $$[N,L]$$ being the composite of $$N$$ and $$L$$, I'd like to show that $$\sigma(N) \subset N$$.

I'm aware of the implication from the Fundamental Theorem of Galois Theory that says that for $$K\leq E\leq F$$ with $$F|K$$ and $$E|K$$ Galois extensions, $$\sigma \in \text{Gal}(F|K): \sigma(E) \subset E$$, which would solve my problem. But I can't seem to show that $$[N,L]|K$$ is Galois (I know an extension being Galois is not transitive). Does it suffice to say that $$[N,L]$$ is also the splitting field over $$K$$ of a separable polynomial, just like $$N$$, although $$[N,L]$$ would eventually contain more elements?

This is my first question here, so let me know if I can improve the format of my question.

• Let $M/K$ be a finite field extension, and $E,F$ two intermediate fields $E,F$. If $E/K$ is Galois, then $EF/F$ is Galois. If $E/K$ and $F/K$ are both Galois, then $EF/K$ is Galois. Feb 24, 2023 at 9:57
• Thank you. Both points are clear to me. So that means, generally, $EF|K$ is not Galois, if for example $F|K$ is not Galois. Do you have an idea how to show the statement in the first sentence of my question without having $EF|K$ be a Galois extension? Feb 24, 2023 at 10:17
• Just to give an easy example of $EF/K$ not generally being Galois: Take a non-Galois extension $L/K$ with $N\subset L$ (you know such examples, because you know that being Galois is not transitive). Then the compositum $NL=L$, so $NL/K$ is not Galois.
– CPCH
Feb 24, 2023 at 15:48

Let $$M/K$$ be a finite field extension, and $$E,F$$ two intermediate fields with $$E/K$$ Galois. Then $$EF/F$$ is Galois and there is an injective group homomorphism $$\mathrm{Gal}(EF/F) \to \mathrm{Gal}(E/K), \quad \sigma \mapsto \sigma|_E$$ To see this, write $$E$$ as the splitting field of a separable polynomial $$f\in K[X]$$. Then $$E/K$$ is generated by the roots of $$f$$, so $$EF=F(E)$$ is generated over $$F$$ by the roots of $$f$$. Thus $$EF/F$$ is the splitting field of $$f\in F[X]$$, which is still separable, and hence $$EF/F$$ is Galois.
Now, if $$\sigma\in\mathrm{Gal}(EF/F)$$, then $$\sigma$$ permutes the roots of $$f$$, so restricts to an automorphism of $$E$$. It is easy to check that we have a group homomorphism between the Galois groups. Finally, if $$\sigma$$ is the identity on $$E$$, then it fixes all the roots of $$f$$, so $$\sigma$$ is the identity on $$EF$$, and the group homomorphism is injective.
• Thank you! Thinking of an element of $\text{Gal}(EF|F)$ as a permutation on the roots of the same separable polynomial as the one whose splitting field over $K$ is $E$ was very helpful! Feb 24, 2023 at 17:10